SOLUTION: Calculating trigonometric equations for x in radians.
Find all the solutions of the equation in the interval [0,2pi).
Recall that the quadrants in standard positions are numbered
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-> SOLUTION: Calculating trigonometric equations for x in radians.
Find all the solutions of the equation in the interval [0,2pi).
Recall that the quadrants in standard positions are numbered
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Question 1175310: Calculating trigonometric equations for x in radians.
Find all the solutions of the equation in the interval [0,2pi).
Recall that the quadrants in standard positions are numbered counterclockwise, starting in the upper right-hand corner.
1.5cosx + Square root 3 = 3cosx
2.3tan²x - 1 = 0
3.2cos²x - cosx = 1
4.4sin³x + 2sin²x - 2sinx - 1 = 0
5.2sinxcosx - sinx = 0
Treat each equation as an equation in which the variable(s) is/are trig functions. Solve by regular algebraic methods; then use your unit circle.
I'll just get you started on each one; you won't learn anything from this if I do all the work for you.
1. 5cosx + Square root 3 = 3cosx
This is a linear equation in cosx. Solve by getting the variable on one side and the number on the other:
2cosx = -sqrt(3)
...
2. 3(tanx)^2 - 1 = 0
This is a quadratic equation in tanx. It is also in the form of a difference of squares. Factor it that way.
(sqrt(3)tanx+1)(sqrt(3)tanx-1) = 0
tanx = -1/sqrt(3) or tanx = 1/sqrt(3)
...
3. 2(cosx)^2 - cosx = 1
This is quadratic in cosx; get everything on one side and factor.
2(cosx)^2-cosx-1 = 0
(2cosx+1)(cosx-1) = 0
...
4. 4(sinx)^3 + 2(sinx)^2 - 2sinx - 1 = 0
A cubic in sinx; 4 terms, so factor by grouping.
(2(sinx)^2)(2sinx+1)-(2sinx+1) = 0
{2(sinx)^2-1)(2sinx+1) = 0
...
5.2sinxcosx - sinx = 0
This is quadratic in both sinx and cosx. Factor out the factor common to both terms.
(sinx)(2cosx-1) = 0
...
